Finding higher order Derivatives

[tmlrlz]

Homework Statement

Let p be an arbitrary polynomial
p(x) = a_nxⁿ + a_n-1x^n-1 + ... + a₁x + a₀, a_n cannot equal 0.
a) Find (dⁿ/dxⁿ)[p(x)]
b)What is (d^k/dx^k)[p(x)] for k>n

Homework Equations

The Attempt at a Solution

Im actually not really sure what to do for this question.

For the first part i tried to take the first three derivatives and got

(d/dx)[p(x)] = a₁
(d²/dx²)[p(x)] = 2a₂x + a₁
(d³/dx³)[p(x)] = 3a₃x² + 2a₂x + a₁

(dⁿ/dxⁿ)[p(x)] = na_nx^n-1 + (n-1)a_(n-1)x^n-2 + ... + a₁

Thats my guess, I am not sure of how to do it.
As for part b:
I don't even know how to approach this question, my only assumption is that its the same answer again with all the n's replaced with k's. I am certain this is wrong so please help me do this question both a and b. Thank you.

 

[I like Serena]

Welcome to PF, tmlrlz!

What you have calculated is d/dx in each case.
dⁿ/dxⁿ means that you take the derivative n times.

 

[tmlrlz]

Welcome to PF, tmlrlz!

What you have calculated is d/dx in each case.
dⁿ/dxⁿ means that you take the derivative n times.

I thought i did that for part a, did i do it wrong? And how would you go about doing part b, what is it asking and how is it different from part a? Thank you.

 

[I like Serena]

I thought i did that for part a, did i do it wrong? And how would you go about doing part b, what is it asking and how is it different from part a? Thank you.

Let me give you an example:

You wrote:
(d²/dx²)[p(x)] = 2a₂x + a₁

But what you actually did was:
(d/dx)[p(x)] = 2a₂x + a₁

It should be:
(d²/dx²)[p(x)] = 2a₂

Can you do this for say n=3?



If you understand what is intended for part a, I think you'll see what is intended for part b.

 

[murmillo]

The difference is that the n in the first part is the degree of the polynomial. The k in the second part is just any integer greater than the degree of the polynomial.

 

[I like Serena]

The difference is that the n in the first part is the degree of the polynomial. The k in the second part is just any integer greater than the degree of the polynomial.

Yes, this is true.
So I guess you answered the last part of your own question.

 

[tmlrlz]

Let me give you an example:

You wrote:
(d²/dx²)[p(x)] = 2a₂x + a₁

But what you actually did was:
(d/dx)[p(x)] = 2a₂x + a₁

It should be:
(d²/dx²)[p(x)] = 2a₂

Can you do this for say n=3?


If you understand what is intended for part a, I think you'll see what is intended for part b.

oh i can't believe that i forgot to do the rest of the derivative!
so for n = 3
(d³/dx³)[p(x)] = 6a₃

And then you can find part a
(dⁿ/dxⁿ)[p(x)] = n! * (a_n)

However i am still confused as to what part b is requiring?

 

[I like Serena]

oh i can't believe that i forgot to do the rest of the derivative!
so for n = 3
(d³/dx³)[p(x)] = 6a₃

And then you can find part a
(dⁿ/dxⁿ)[p(x)] = n! * (a_n)

However i am still confused as to what part b is requiring?

Right!

Okay, let's pick a k>n.
I'll pick k=n+1.

What is (dⁿ⁺¹/dxⁿ⁺¹)[p(x)]?

 

[tmlrlz]

Right!

Okay, let's pick a k>n.
I'll pick k=n+1.

What is (dⁿ⁺¹/dxⁿ⁺¹)[p(x)]?

So then that means that when n = 1 we are taking the derivative of d²/dx² right?

I took the first 3 derivatives again
when n =1
d²/dx²[p(x)] = d/dx (a₁) = 0
when n =2
d³/dx³[p(x)] = 0
when n =3
d⁴/dx⁴[p(x)] = 0

so that means that
(d^k/dx^k)[p(x)] = 0

right? i hope!

 

[I like Serena]

So then that means that when n = 1 we are taking the derivative of d²/dx² right?

I took the first 3 derivatives again
when n =1
d²/dx²[p(x)] = d/dx (a₁) = 0
when n =2
d³/dx³[p(x)] = 0
when n =3
d⁴/dx⁴[p(x)] = 0

so that means that
(d^k/dx^k)[p(x)] = 0

right? i hope!

Yep!

What do you think will happen if we pick k=n+2?
Or k=n+3?
Or any k>n?

 

[tmlrlz]

Yep!

What do you think will happen if we pick k=n+2?
Or k=n+3?
Or any k>n?

Then you will keep taking the derivative of zero which is zero over and over again till the end of the derivative instead of the previous case where you only got zero once, instead it will occur multiple times for that specific derivative

 

[I like Serena]

So? Any other questions?

 

[tmlrlz]

So? Any other questions?

Thank you so much for your help. I do have more questions but i do not wish to trouble you, if you would like you can look at the question i posted, i seem to be having trouble on it even though people have been trying to help me, https://www.physicsforums.com/showthread.php?p=3593450&posted=1#post3593450
Thank you once again!